Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int IN() {
  int c, f, x;
  while (!isdigit(c = getchar()) && c != '-')
    ;
  c == '-' ? (f = 1, x = 0) : (f = 0, x = c - '0');
  while (isdigit(c = getchar())) x = (x << 1) + (x << 3) + c - '0';
  return !f ? x : -x;
}
const int N = 100 + 19;
const int oo = (1 << 30) - 1;
int A[N][N], B[N][N], c[N];
int n, m, rev;
vector<int> V;
int main() {
  n = IN(), m = IN();
  for (int i = 1; i < n + 1; i++)
    for (int j = 1; j < m + 1; j++) A[i][j] = IN();
  if (n > m) {
    swap(n, m);
    for (int i = 1; i < n + 1; i++)
      for (int j = 1; j < m + 1; j++) B[i][j] = A[j][i];
    for (int i = 1; i < n + 1; i++)
      for (int j = 1; j < m + 1; j++) A[i][j] = B[i][j];
    rev = 1;
  }
  for (int i = 1; i < n + 1; i++) {
    int mn = oo;
    for (int j = 1; j < m + 1; j++) mn = min(mn, A[i][j]);
    for (int j = 1; j < m + 1; j++) A[i][j] -= mn;
    for (int j = 0; j < mn; j++) V.push_back(i);
  }
  for (int j = 1; j < m + 1; j++) {
    int chk = 1;
    for (int i = 1; i < n + 1; i++)
      if (A[i][j] != A[1][j]) chk = 0;
    if (!chk) {
      puts("-1");
      return 0;
    }
    for (int i = 0; i < A[1][j]; i++) V.push_back(-j);
  }
  printf("%d\n", V.size());
  for (int i = 0; i < V.size(); i++) {
    if ((V[i] > 0) == (rev == 0))
      printf("row ");
    else
      printf("col ");
    printf("%d\n", abs(V[i]));
  }
}